Potential Due to Dipole

Description:

We know that two equal and opposite charges fixed at a finite distance makes a Dipole. This dipole is able to create an electric field, so there will be potential linked at every point in the electric field.

To calculate potential created by a dipole on the axial line, take a dipole having charges –q and +q.

• All the measurement of distances are to be taken from the centre(O).

• Let the distance between O to +q and O to –q be ‘l’. So, total length between +q and –q will be ‘2l’.

• Take a point ‘p’ on the axial line at the distance ‘r’ from the centre as shown in figure.

Now, we wish to calculate potential at point ‘P’

By using the formula for potential due to point charge,

potential due to +q = + 1/4πε₀ q/(r-1)

The distance between (P and +q) = (r-l)

potential due to -q = - 1/4πε₀ q/(r-1)

The distance between (P and -q) = (r+l)

(potential due to +q will be positive and potential due to- q will be negative)

Since potential is a scalar quantity so, the net potential will be the scalar addition of the two.

So, the net potential V = V₁ + V₂

V = +1/4πε₀ q/(r-1) - 1/4πε₀ q/(r+1)

V = q/4πε₀[1/(r-1) - 1/(r+1)]

On solving the equation we get

V = q/4πε₀[(r+1)-(r-1)/(r²-1²)]

V = q/4πε₀ 2l/(r²-1²).....(1)

We know that the dipole moment or effectiveness of dipole (P) is given by −

P = 2ql

Therefore, putting this value in eq(1), we get

V 1/4πε₀ P/(r² - 1²).....(2)

certain assumptions are made based on this equation:

since, the dipole is very small so ‘l’ is also very small as compared to the distance ‘r’.

So, on neglecting ‘l’ with respect to ‘l’ we get:

V = 1/4πε₀ P/(r²) .....(from eq(2))